# Composing covers with epis

I am beginner of sheaf-theory and beg your pardon for this maybe silly question.

Let $\mathcal{C}$ be a Grothendieck site and $T$ the category of sheaves on $\mathcal{C}$ and let $f:X\rightarrow Y$ be an epic morphism in $T$ into a representable sheaf $Y$.

I have a general lack of understanding how such epic morphisms look like and this leads to the the questions:

• Suppose $\{Y_\alpha\to Y\}$ is a cover of $Y$. Is $Y_\alpha\times_Y X$ representable? I see no formal reason why this is true but somehow my miserable intuition still thinks it might be.
• (If the answer to the first point is ''no'', please assume a representable $X$) Suppose $\{X_\alpha\to X\}$ is a cover of $X$. Is the composition $\{X_\alpha\to X\xrightarrow{f} Y\}$ a cover of $Y$?
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The fibre product of two representable sheaves over a representable sheaf is itself representable... if the site itself has those fibre products. –  Zhen Lin May 16 '12 at 15:52

In more detail.

1. By definition, limits commute with hom: $$\mathcal{C}(-, \varprojlim x_\bullet) \cong \varprojlim \mathcal{C} (-, x_\bullet)$$ In particular, fibre products commute with hom: $$\mathcal{C}(-, x \times_z y) \cong \mathcal{C}(-, x) \times_{\mathcal{C}(-, z)} \mathcal{C}(-, y)$$ So far this has nothing to do with Grothendieck topologies. If you have a subcanonical Grothendieck topology then all this still applies verbatim; otherwise you have to replace $\mathcal{C}(-, x)$ with its sheafification, and the claim only holds for finite limits because sheafification is only left exact.

Now, if $Y$ and $Z$ are representable but $X$ is not, then it does not follow that $X \times_Z Y$ is representable. For example, $\mathcal{C}$ could be a site with finitely many objects and finitely many morphisms, while $X$ (and hence $X \times_Z Y$ could be infinite.)

2. I presume by ‘cover’ you mean jointly epimorphic. The answer is yes. In a Grothendieck topos, all small coproducts exist, so you can replace a jointly epimorphic family $\{ X_\alpha \to X \}$ with a single epimorphism $\coprod_\alpha X_\alpha \to X$. It is a fact about general categories that the latter morphism, when it exists, is epimorphic if and only if the original family is jointly epimorphic. The composition of two epimorphisms is obviously an epimorphism, hence, $\{ X_\alpha \to X \to Y \}$ is jointly epimorphic if $X \to Y$ is an epimorphism.

The question is slightly more subtle if we work in the underlying site. There, one cannot take coproducts (even if they exist) and one must look at the Grothendieck topology $J$ itself. One of the axioms of Grothendieck topologies is that covers compose: if $\mathfrak{U}$ is a sieve on $Y$ such that the pullback sive $f^* \mathfrak{U}$ is a $J$-covering sieve on $U$ for each $f : U \to Y$ in $\mathfrak{U}$, then $\mathfrak{U}$ is also a $J$-covering sieve. In terms of a Grothendieck pretopology $K$, if $\{ U_\beta \to Y \}$ is a $K$-covering family and $\{ X_{\alpha, \beta} \to U_\beta \}$ is a $K$-covering family for each $U_\beta$, then the composite family $\{ X_{\alpha, \beta} \to U_\beta \to Y \}$ is also a $K$-covering family.

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I need help with your second answer. In my situation $f:X\rightarrow Y$ is a morphism of the site $\mathcal{C}$ such that the Yoneda embedded $\tilde f:\tilde X\rightarrow\tilde Y$ is en epi of sheaves. If now $\{X_\alpha\rightarrow X\}$ is a sieve on $X$, is the composition $\{X_\alpha\rightarrow X\rightarrow Y\}$ a sieve on $Y$? My problem is then the relation between epis (or jointly epimorphic families) in the site $\mathcal{C}$ and epis (or jointly epimorphic families) in the category of sheaves on $\mathcal{C}$. How are they related? –  Shinze May 18 '12 at 7:46
A family of morphisms generates a $J$-covering sieve in the site if and only if it becomes jointly epimorphic in the sheaf topos. ($J$ really has to be a Grothendieck topology for this to be true.) –  Zhen Lin May 18 '12 at 7:51
Wow, this would be perfect. I think, I can show that if $\{X_\alpha\rightarrow X\}$ is a cover, then it becomes jointly epimorphic in the sheaf topos. Do you know how I can show the converse? –  Shinze May 18 '12 at 8:04
The joint presheaf image of $\{ \tilde{X}_\alpha \to \tilde{X} \}$ is a presheaf subobject of $\tilde{X}$, and there is a bijective correspondence between presheaf subobjects of $\tilde{X}$ and (not necessarily covering) sieves on $X$. If the joint sheaf image is the whole of $\tilde{X}$, then the joint presheaf image is dense, and a dense presheaf subobject of $\tilde{X}$ is exactly a $J$-covering sieve. –  Zhen Lin May 18 '12 at 8:08
I think I understand it better now, thanks. Doesn't this mean, that when I take a Grothendieck pretopology $K$ which is maximal in the sense that it can't be refined anymore without changing the associated Grothendieck topos, then a family $\{\tilde X_\alpha\rightarrow \tilde X\}$ of representables is jointly epimorphic iff $\{X_\alpha\rightarrow X\}$ is a cover of $K$? –  Shinze May 21 '12 at 13:30